Computer Science, asked by kishoreduggemi, 9 months ago

Here on earth, our 24-hour day is composed of two parts, each of 12 hours. Each hour in each part has a corresponding hour in the other part separated by 12 hours: the hour essentially measures the duration since the start of the day part. For example, 1 hour in the first part of the day is equivalent to 13, which is 1 hour into the second part of the day.

Answers

Answered by TCS2006
2

Answer:

Where is the question?

Answered by serah99john
0

Problem Description

Here on earth, our 24-hour day is composed of two parts, each of 12 hours. Each hour in each part has a corresponding hour in the other part separated by 12 hours: the hour essentially measures the duration since the start of the day part. For example, 1 hour in the first part of the day is equivalent to 13, which is 1 hour into the second part of the day.

Now, consider the equivalent hours that are both prime numbers. We have 3 such instances for a 24-hour 2-part day:

5~17

7~19

11~23

Accept two natural numbers D, P >1 corresponding respectively to number of hours per day and number of parts in a day separated by a space. D should be divisible by P, meaning that the number of hours per part (D/P) should be a natural number. Calculate the number of instances of equivalent prime hours. Output zero if there is no such instance. Note that we require each equivalent hour in each part in a day to be a prime number.

Example:

Input: 24 2

Output: 3 (We have 3 instances of equivalent prime hours: 5~17, 7~19 and 11~23.)

Constraints

10 <= D < 500

2 <= P < 50

Input

Single line consists of two space separated integers, D and P corresponding to number of hours per day and number of parts in a day respectively

Output

Output must be a single number, corresponding to the number of instances of equivalent prime number, as described above

Time Limit

1

Examples

Example 1

Input

36 3

Output

2

Explanation

In the given test case D = 36 and P = 3

Duration of each day part = 12

2~14~X

3~15~X

5~17~29 - instance of equivalent prime hours

7~19~31 - instance of equivalent prime hours

11~23~X

Hence the answers is 2.

Example 2

Input

49 7

Output

0

Explanation

Duration of each day part = 7

2~9~X~23~X~37~X

3~X~17~X~31~X~X

5~X~19~X~X~X~47

7~X~X~X~X~X~X

Hence there are no equivalent prime hours

Answer:

#include<stdio.h>

int count=0;

int prime(int a)  //to check whether prime or not

{

 int flag=0;

 for(int j=2; j<=(a/2);j++)

       {

          if(a%j==0)

          {

             flag=1;

             break;

           }

       }

       

 return flag;

}

int find(int b,int c,int d,int ec)

{

   

 int arr[10],ch=0;

 for(int i=0;i<d;i++)

 {

   arr[i]=prime(b);

   b=b+c;

   if(b>ec)

   {

       break;

   }

     }

  for(int i=0;i<d;i++)

  {

       

    if(arr[i]==0)

      ch++;

       

  }

 if(ch==d)

 return 1;

   

 else

   

     return 0;

}

   

int main()

{

 int D;

 int P;

 scanf("%d",&D);

 scanf("%d",&P);

 if(D%P==0)

 {

   int q=D/P;

   

     for(int i=2;i<=D;i++)

   {

       count=count+find(i,q,P,D);

       

         

             

   }

   printf("%d",count);

   

 }

 return 0;

}

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